Quantitative Aptitude - Arithmetic Ability Questions

A can do the work = 18 days
B can do the work = 18/2 = 9 days
(A + B)'s 1 day work = 1/18 + 1/9 = 1/6
=> In 3 days = 3x1/6 = 1/2 work is completed.

(A+B)'s 1 day's work=1/20 

C's 1 day work=1/60 

(A+B+C)'s 1 day's work= 1/20 + 1/60 = 1/15

Also A's 1 day's work =(B+C)'s 1 day's work 

Therefore,  we get: 2 x (A's 1 day 's work)=1/15 

=>A's 1 day's work=1/30 

Therefore, B's 1 day's work= 1/20 - 1/30 = 1/60

So, B alone could do the work in 60 days.

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 2024, Which is a leap year.

So, Add +28 to the given year (i.e 2024 + 28) = 2052

Therfore, The calendar of the year 2024 can be used again in the year 2052.

Let the sum be Rs. P.

Then,[p(1+10/100)2-p]=525

Sum =Rs.2500

S.I.= Rs.(2500*5*4)/100

= Rs. 500

Age of the 15th student = [15 * 15 - (14 * 5 + 16 * 9)] = (225-214) = 11 years.

Let their prices be 3x,  5x and 7x.

Then, 3x + 5x + 7x = (15000 * 3) or x = 3000.

Cost of cheapest item = 3x = Rs. 9000.